How much force is needed to stop a 50 kg gymnast if he decelerates at 25 m/s^2
1 answer:
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Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV =
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p = 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² =
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c =
v/c= 2.33 10⁻⁴
Answer:
a)
125.6 rad/s
b)
25.12 rad/s²
Explanation:
a)
t = time required by the fan to get up to final operating speed = 5 sec
w = final operating rotational speed = 1200 rpm
we know that :
1 revolution = 2π rad
1 min = 60 sec
w =
w =
w = 125.6 rad/s
b)
w₀ = initial angular speed = 0 rad/s
α = angular acceleration
using the equation
w = w₀ + α t
125.6 = 0 + α (5)
α = 25.12 rad/s²